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Proof step 15 from proof diffgetn-nodupssize (previous step)

m0 < # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ a = x0[m0], #(x0 \ x0[m0] ') + 1 = # x0, x0 = [], a ∈ x0, dups(x0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: # x0 - 1, m0 abort: F discard: T computed: F

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat Used terms: # x0 - 1, m0